I have here a linear transformation $T : P_3(\mathbb{R})\rightarrow P_3(\mathbb{R}) $ defined by:
$ T(at^3 + bt^2 + ct + d) = (a-b)t^3 + (c-d)t $
I'm very very new in this subject and I'm not going well with polynomials. I need find the $ Kernel $ and the $ Image $ of the transformation. Look what I've been thinking:
$Ker(T) = \{ T(p) = 0 / p \in P_3\} $
$ T(at^3 + bt^2 + ct + d) = (a-b)t^3 + (c-d)t = 0 $
$(a-b) = 0 \ ;\ \ (c-d) = 0 \ ;\ \ a = b \ ; \ \ c = d $
$ Ker(T) = \{ at^3 + at^2 + ct +c\ /\ a,c \in \mathbb{R} \} $
And what about the $ Image $? I know that $Im(T) = \{ T(p) / p \in P_3 \}$, but how can I show it? And how can I test if a polynomial such as $ p(t) = t^3 + t^2 + t -1 \in Im(t)$?
Best Answer
The kernel is correct. Additionally, since the kernel depends on only two coefficients $a$ and $c$, it has dimension 2.
For the image:
Take any polynomial $p(t)=At^3+Bt^2+Ct+E$.
The question now is: How do $A,B,C,E$ have to look for there to exist some $a,b,c,d$ such that $T(at^3+bt^2+ct+d)=p(t)$?
The question is equivalent to solving for $A, B, C, E$ in the equation: $(a-b)t^3+0t^2+(c-d)t+0=At^3+Bt^2+Ct+E$.
We now have:
$A=a-b$,
$B=0$,
$C=c-d$
$E=0$
We can take:
$a=A$,
$b=0$,
$c=C$
$d=0$
Consequently, $p$ is in the image, iff $B=0=E$. The image, then, is: \begin{align*} \mbox{Im}(T)=\{At^3+Ct\ |\ A,C\in\mathbb R\}. \end{align*}